MAT334-2018F > Quiz-3

Q3 TUT 5101

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**Victor Ivrii**:

Let $D$ be the domain obtained by deleting the ray $\{x\colon x \le 0\}$ from the plane, and let $G(z)$ be a branch of $\log z$ on $D$. Show that $G$ maps $D$ onto a horizontal strip of width of $\pi$,

$$

\{x + iy\colon - \infty < x < \infty,\ c_0 < y < c_0 + 2\pi\},

$$

and that the mapping is one-to-one on $D$.

Draw both domains.$\newcommand{\Arg}{\operatorname{Arg}}$

**Kathy Ngo**:

Fix $G(z)$ such that $G(z)=\ln|z|+i(Arg(z) + c_{0} +\pi)$ where $c_{0} \in \mathbb{R}$

let $x=\ln|z|$ and $y=(Arg(z) + c_{0} +\pi)$

Note that

$\forall z\in D, |z|\geq 0 \Rightarrow x=\ln|z|\in(-\infty, \infty)$ and

$\forall z\in D, Arg(z) \in (-\pi, \pi) \Rightarrow y=Arg(z) + c_{0} +\pi \in (c_{0} , c_{0}+2\pi)$Therefore $G$ maps $D$ onto $\{x+iy: -\infty < x< \infty, c_{0} <y<c_{0}+2\pi\}$.

Suppose $G(z_{1}) = G(z_{2})$ then

$\ln|z_{1}|+i(Arg(z_{1}) + c_{0} +\pi)=\ln|z_{2}|+i(Arg(z_{2}) + c_{0} +\pi)$therefore, we have

$\ln|z_{1}| = \ln|z_{2}|$

$Arg(z_{1}) + c_{0} +\pi = Arg(z_{2}) + c_{0} +\pi \Rightarrow Arg(z_{1})=Arg(z_{2})$ these two equations imply $z_{1} = z_{2}$, hence the mapping is one-to-one on $D$.

**Victor Ivrii**:

Arguably you need to type Arg upright and provide a proper spacing after it. The trouble is that LaTeX (and Mathjax) would not recognize \Arg since it was not defined. But one can use \operatorname{Arg} ; however one could do better then this defining (only once!) \Arg

--- Code: ---\newcommand{\Arg}{\operatorname{Arg}}

--- End code ---

(I did it).

Similarly, one can change existing definition

--- Code: ---\renewcommand{\Re}{\operatorname{Re}}

--- End code ---

On the pictures do not use xxxxxx for "cuts". Just straight line

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